Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 7 - Rational Functions - 7.4 Adding and Subtracting Rational Expressions - 7.4 Exercises: 2

Answer

$\text{LCD: } 75ab^2c^3 \\\\\text{Equivalent of 1st Expression: } \dfrac{55a^2c^2}{75ab^2c^3} \\\\\text{Equivalent of 2nd Expression: } \dfrac{42b^3}{75ab^2c^3}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the $LCD$ of the given expressions, $ \dfrac{11a}{15b^2c} $ and $ \dfrac{14b}{25ac^3} ,$ express the denominators in factored form. Then take the highest exponent of each factor. Finally use the $LCD$ to find the required equivalent expression for the given. $\bf{\text{Solution Details:}}$ In factored form, the given expressions are equivalent to \begin{array}{l}\require{cancel} \dfrac{11a}{3(5)(b^2)(c)} \text{ and } \dfrac{14b}{(5^2)(a)(c^3)} .\end{array} Taking the highest exponent of each factor, then the $LCD$ is \begin{array}{l}\require{cancel} 3(5^2)(a)(b^2)(c^3) \\\\= 3(25)(a)(b^2)(c^3) \\\\= 75ab^2c^3 .\end{array} Multiplying the given expression by an expression equal to $1$ such that the denominator becomes the $LCD,$ then the given expressions are equivalent to \begin{array}{l}\require{cancel} \dfrac{11a}{15b^2c} \cdot\dfrac{5ac^2}{5ac^2}= \dfrac{55a^2c^2}{75ab^2c^3} \\\\\text{ and }\\\\ \dfrac{14b}{25ac^3} \cdot\dfrac{3b^2}{3b^2}= \dfrac{42b^3}{75ab^2c^3} .\end{array} Hence, the needed pieces of information are as follows: \begin{array}{l}\require{cancel} \text{LCD: } 75ab^2c^3 \\\\\text{Equivalent of 1st Expression: } \dfrac{55a^2c^2}{75ab^2c^3} \\\\\text{Equivalent of 2nd Expression: } \dfrac{42b^3}{75ab^2c^3} .\end{array}
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